$M g S O_{4}$ on reaction with $N H_{4} O H$ and $N a_{2} H P O_{4}$ forms a white crystalline precipitate. What is its formula?

$M g (N H_{4}) P O_{4}$

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$M g_{3} P O_{4}$

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$M g C l_{2} . M g S O_{4}$

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$M g S O_{4}$

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Solution

The correct option is A
$M g (N H_{4}) P O_{4}$

The white crystalline precipitate is Ammonium magnesium phosphate:

$M g S O_{4} + N a_{2} H P O_{4} + N H_{4} O H ⟶ M g (N H_{4}) P O_{4} + N a_{2} S O_{4} + H_{2} O$

Suggest Corrections

Similar questions

M g S O_{4}

N H_{4} O H

N a_{2} H P O_{4}

M g S O_{4}

N H_{4} C I

N a_{2} H P O_{4}

M g S O_{4} + N H_{4} O H + N a_{2} H P O_{4} \to

M g S O_{4}

N H_{4} C l

N a_{3} P O_{4}

B e S O_{4}, M g S O_{4}, C a S O_{4}

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